SUPPRESSION OF VIBRATION IN STRETCHED STRINGS BY THE BOUNDARY CONTROL

Journal of Sound and Vibration (1997) 204(5), 835–840

SUPPRESSION OF VIBRATION IN STRETCHED STRINGS BY THE BOUNDARY CONTROL S. M. S  C. A. N Berkeley Engineering Research Institute, P.O. Box 9984, Berkeley, California 94709, U.S.A. (Received 29 January 1997, and in final form 10 February 1997)

1.  In this note, an elastic string of unit length and unit mass density, represented by the following non-linear partial differential equation, is considered ytt (x, t) = [1 + 32 byx2 (x, t)]yxx (x, t),

(1a)

for all x $(0, 1) and t e 0, with the boundary conditions y(0, t) = 0,

T(1, t)yx (1, t) = u(t),

(1b, c)

for all t e 0, and the initial conditions y(x, 0) = f(x),

yt (x, 0) = g(x),

(1d)

for all x $(0, 1). In equations (1), y(·, ·)$R denotes the transversal displacement of the string, T(1, t) q 0 denotes the tension in the string at x = 1 for all t e 0, u(·)$R is a control input, yx M1y/1x, yxx M1 2y/1x 2, yt M1y/1t, ytt M1 2y/1t 2, and b q 0 is a constant real number. There are several non-linear mathematical models that describe the transversal vibration of stretched strings. One such model is presented in equation (1a). This model was derived in reference [1] and has been studied by researchers from the physical and mathematical points of view; see, e.g., references [2–6] and the references therein. The boundary condition in equation (1b) implies that the string is fixed at x = 0. The boundary condition in equation (1c) represents the balance of the transversal component of the tension in the string and the control input u, which is applied transversally at x = 1. The tension in the string represented by equation (1a) is not constant and is given by T(x, t) = 1 + 12 byx2 (x, t),

(2)

for all x $[0, 1] and t e 0 (see references [1, 2]). Therefore, the boundary condition in equation (1c) can be written as [1 + 12 byx2 (1, t)]yx (1, t) = u(t),

(3)

for all t e 0. In equation (1d), the initial displacement and velocity of the string are, respectively, denoted by f(x) and g(x) for all x $(0, 1). One assumes that f $C1[0, 1], and that at least one of the functions f or g is not identically zero over [0, 1]. The control input u in equation (3) is commonly known as the boundary control. In this note, the stabilization of the string in equaion (1a) by u is studied. More precisely, a u that results in y(x, t):0 as t:a for all x $[0, 1], is studied. As a stabilizing control input, one proposes u(t) = −kyt (1, t), 0022–460X/97/300835 + 06 $25.00/0/sv970985

(4) 7 1997 Academic Press Limited

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for all t e 0, where k q 0 is a constant real number. With this choice of u, the boundary control is the negative feedback of the transversal velocity of the string at x = 1, with the gain k. It is known that linear strings represented by equations (1)–(3), for which b = 0, can be stabilized by the control law in equation (4), see, e.g., references [7–12]. Roughly speaking, the boundary control in equation (4) provides a dissipative effect in linear strings, because it is of the form of negative velocity feedback. This is in accordance with the well known fact that the negative velocity feedback increases damping in most finite dimensional inertial systems, such as large flexible systems and robotic manipulators. The authors’ goal in this note is to show that the boundary control u in equation (4) stabilizes the non-linear string in equations (1)–(3), i.e., u results in y(x, t):0 as t:a for all x $[0, 1]. The approach taken in this note to show the stabilization of the string in equations (1)–(4) is similar to that in reference [13], where the stabilization of the Kirchhoff’s non-linear string by the boundary control was presented. 2.     The authors’ plan to establish the stability of the non-linear string represented by equations (1)–(4) is as follows. One defines an energy like (Lyapunov) function of time for the string and denote it by t  V(t). One shows that V tends to zero exponentially. The scalar-valued function V is defined as V(t)ME(t) + g

g

1

xyt (x, t)yx (x, t) dx,

(5)

0

for all t e 0, where g q 0 is a constant real number, E(t)M12

g

1

[ yt2 (x, t) + yx2 (x, t)] dx +

0

b 8

g

1

yx4 (x, t) dx,

(6)

0

and y(·, ·) satisfies equations (1)–(4). From equations (5), (6), and (1d), one obtains E(0) = 12

g

1

[g 2(x) + f x2 (x)] dx +

0

V(0) = E(0) + g

g

b 8

g

1

f x4 (x) dx,

(7a)

0

1

xg(x)fx (x) dx,

(7b)

0

where fx (x)Mdf(x)/dx. Since at least one of the functions f or g is not identically zero over [0, 1], one has E(0) q 0. Now, a property of V is proved. Lemma 2.1. Let g in equation (5) satisfy g Q 1.

(8)

0 E K1 E(t) E V(t) E K2 E(t),

(9)

Then, the function V satisfies for all t e 0, where K1 q 0 and K2 q 0 are constant real numbers given by K1 = 1 − g,

K2 = 1 + g.

(10a, b)

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Proof. For the integral term in equation (5), whose coefficient is g, one has (the argument (x, t) of the functions is deleted)

g

1

xyt yx dx E

0

g

1

x=yt ==yx = dx E 12

0

g

1

yt2 dx + 12

0

g

1

yx2 dx E E(t),

(11)

0

for all t e 0. Similarly, one obtains

g

1

xyt yx dx e −E(t),

(12)

0

for all t e 0. Using equations (11) and (12) in equation (5), one obtains inequality (9). q Remark. Let g satisfy inequality (8). Then, by inequality (9) and the fact that E(0) q 0, it is concluded that V(0) q 0. q Next, equation (4) is used in equation (3) and the boundary conditions are rewritten as y(0, t) = 0,

yx (1, t) = −kyt (1, t)/(1 + byx2 (1, t)/2),

(13a, b)

for all t e 0. One now proves some identities for the functions satisfying equations (13). Lemma 2.2. Let y(·, ·) satisfy the boundary conditions in equations (13). Then,

g

1

kyt2 (1, t) ( yxx yt + yxt yx ) dx = − , 1 + byx2 (1, t)/2

(14a)

k 3yt4 (1, t) (3yxx yx2 yt + yx3 yxt ) dx = − , [1 + byx2 (1, t)/2]3

(14b)

0

g

1

0

g

1

xyxt yt dx = 12 yt2 (1, t) − 12

0

g

1

g

1

yt2 dx,

(14c)

0

xyxx yx dx =

0

xyxx yx3 dx =

0

g

1

k 2yt2 (1, t) −1 2[1 + byx2 (1, t)/2]2 2

g

1

k 4yt4 (1, t) −1 4[1 + byx2 (1, t)/2]4 4

g

1

yx2 dx,

(14d)

yx4 dx,

(14e)

0

0

for all t e 0. Proof. From equation (13a), one has yt (0, t) = 0 for all t e 0. Thus, one obtains

g

1

0

( yxx yt + yxt yx ) dx =

g

1

( yx yt )x dx = yx (1, t)yt (1, t),

0

for all t e 0. Using equation (13b) in equation (15), one obtains equation (14a).

(15)

   

838

Having yt (0, t) = 0 for all t e 0, one next obtains

g

1

(3yxx yx2 yt + yx3 yxt ) dx =

0

g

1

( yx3 yt )x dx = yx3 (1, t)yt (1, t),

(16)

0

for all t e 0. Using equation (13b) in equation (16), one obtains equation (14b). Next one writes

g

1

xyxt yt dx = 12

0

g

g

1

(xyt2 )x dx − 12

0

1

yt2 dx,

(17)

0

for all t e 0. Thus, equation (14c) follows. Next, one writes

g

1

xyxx yx dx = 12

0

g

1

(xyx2 )x dx − 12

0

g

1

yx2 dx = 12 yx2 (1, t) − 12

0

g

1

yx2 dx,

(18)

0

for all t e 0. Using equation (13b) in equation (18), one obtains equation (14d). Finally, one writes

g

1

xyxx yx3 dx = 14

0

g

1

(xyx4 )x dx − 14

0

g

1

yx4 dx = 14 yx4 (1, t) − 14

0

g

1

yx4 dx,

(19)

0

for all t e 0. Using equation (13b) in equation (19), one obtains equation (14e).

q

Next, the time-derivative of the function E is computed. Lemma 2.3. The time-derivative of the function E in equation (6), along the solution of the system (1a), (1d), and (13) (equivalently, the system (1)–(4)) satisfies E (t) = −kyt2 (1, t) E 0,

(20)

for all t e 0. Proof. From equation (6), one obtains E (t) =

g

1

( ytt yt + yxt yx ) dx +

0

b 2

g

1

yxt yx3 dx,

(21)

0

for all t e 0. Substituting ytt from equation (1a) into equation (21), one obtains E (t) =

g

1

( yxx yt + yxt yx ) dx +

0

b 2

g

1

(3yxx yx2 yt + yx3 yxt ) dx,

(22)

0

for all t e 0. Using equation (14a) and (14b) in equation (22), one obtains E (t) = −

$

%

kyt2 (1, t) bk 2yt2 (1, t) 1+ , 2 1 + byx (1, t)/2 2[1 + byx2 (1, t)/2]2

(23)

for all t e 0. Using equation (13b) in the last term of equation (23), one obtains equation (20). q

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Using the preliminary results obtained thus far, it is next proved that the functions V and E tend to zero exponentially. Theorem 2.4. Let g in equation (5) satisfy g Q 4k/(3k2 + 2).

(24)

Then, the functions V and E, along the solution of the system (1a), (1d), and (13) (equivalently, the system (1)–(4)) satisfy 0 E V(t) E V(0) e−(g/K2)t

0 E E(t) E (V(0)/K1 ) e−(g/K2)t,

(25a, b)

for all t e 0, where K1 and K2 are given in equations (10). Proof. From equation (5), one obtains V (t) = E (t) + g

g

1

(xytt yx + xyt yxt ) dx,

(26)

0

for all t e 0. Substituting ytt from equation (1a) into equation (26), one obtains V (t) = E (t) + g

g

1

(xyxt yt + xyxx yx + 32 bxyxx yx3 ) dx,

(27)

0

for all t e 0. Using equations (20), (14c), (14d), and (14e) in equation (27), one obtains V (t) = −gE(t) −

gb 4

g

1

0

g gk 2yt2 (1, t) yx4 (x, t) dx − kyt2 (1, t) + yt2 (1, t) − 4[1 + byx2 (1, t)/2]2 2

$

%

3gk 2yt2 (1, t) bk 2yt2 (1, t) + , 2 2 1+ 4[1 + byx (1, t)/2] 2[1 + byx2 (1, t)/2]2

(28)

for all t e 0. Neglecting the second and fifth terms of equation (28) and using equation (13b) in the last term of this equation, one obtains g 3gk 2yt2 (1, t) , V (t) E −gE(t) − kyt2 (1, t) + yt2 (1, t) + 2 4[1 + byx2 (1, t)/2]

(29)

for all t e 0. Therefore, V (t) E −gE(t) − F(t),

(30)

F(t)M[k − g(3k 2 + 2)/4]yt2 (1, t).

(31)

for all t e 0, where Having inequality (24), one concludes that F(t) e 0 for all t e 0. Using the non-negativeness of F in inequality (30), one obtains V (t) E −gE(t),

(32)

for all t e 0. Since 4k/(3k + 2) E z(2/3) Q 1 for all k q 0, from inequality (24), one concludes that g Q 1. Therefore, inequalities (8) and (9) hold. Using inequality (9) in inequality (32), one obtains the differential inequality 2

V (t) E −(g/K2 )V(t),

(33)

840

   

for all t e 0, with the initial condition V(0) q 0 given in equation (7b). By a comparison theorem given in references [14, p. 2] or [15, p. 3], one concludes that V in inequality (33) satisfies V(t) E V(0) e−(g/K2)t for all t e 0. Note that by inequality (9), one has V(t) e 0 for all t e 0. Thus, inequality (25a) holds. By inequalities (9) and (25a), one concludes that inequality (25b) holds. q Finally, it is shown that the boundary control u in equation (4) stabilizes the non-linear string in equations (1)–(3). Corollary 2.5. The solution of the system (1a), (1d), and (13) (equivalently, the system (1)–(4)) y(x, t):0 as t:a for all x $[0, 1]. Proof. For the system (1a), (1d), and (13), one chooses the Lyapunov function V in equation (5), and lets g in equation (5) satisfy inequality (24). Then, by Theorem 2.4, the function E tends to zero exponentially. From equation (6), one concludes that yx (x, t):0 as t:a for all x $[0, 1]. Since y(0, t) = 0 for all t e 0, one concludes that y(x, t):0 as t:a for all x $[0, 1]. q 3.  In this note, it was proved that the non-linear stretched string represented by equations (1)–(3) can be stabilized by the linear boundary control in equation (4). The boundary control is the negative feedback of the transversal velocity of the string at one end.  1. E. L. L 1957 British Journal of Applied Physics 8, 411–413. Non-linear forced vibration of a stretched string. 2. G. S. S. M and B. S. R 1965 Journal of the Acoustical Society of America 38, 461–471. Nonlinear character of resonance in stretched strings. 3. G. V. A 1966 Journal of the Acoustical Society of America 40, 1517–1528. Nonlinear resonance in stretched strings with viscous damping. 4. G. V. A and K. R 1974 International Journal of Non-Linear Mechanics 9, 251–260. Non-linear response of a string to random excitation. 5. A. A and A. B 1989 Physical Reviews Letters 63, 1230–1232. Nonlinear strings as bistable elements in acoustic wave propagation. 6. M. C and R. K 1992 Quarterly of Applied Mathematics 50, 57–71. Forced oscillations of elastic strings with nonlinear damping. 7. G. C 1979 Journal de Mathe´matiques Pures et Applique´es 58, 249–273. Energy decay estimates and exact boundary value controllability for the wave equation in a bounded domain. 8. G. C 1981 SIAM Journal of Control and Optimization 19, 106–113. A note on boundary stabilization of the wave equation. 9. V. K and E. Z 1990 Journal de Mathe´matiques Pures et Applique´es 69, 33–54. A direct method for the boundary stabilization of the wave equation. 10. J. L 1983 Journal of Differential Equations 50, 163–182. Decay of solutions of wave equations in a bounded region with boundary dissipation. 11. J. E. L 1988 SIAM Journal of Control and Optimization 26, 1250–1256. Note on boundary stabilization of wave equations. 12. J. Q and D. L. R 1977 Proceedings of the Royal Society of Edinburgh 77A, 97–127. Asymptotic stability and energy decay rates for solutions of hyperbolic equations with boundary damping. 13. S. M. S and L. G. K 1996 Journal of Sound and Vibration 195, 169–174. Boundary control of a non-linear string. 14. D. B and P. S 1992 Integral Inequalities and Applications. Dordrecht, The Netherlands: Kluwer Academic Publishers. 15. V. L, S. L and A. A. M 1989 Stability Analysis of Nonlinear Systems. New York: Marcel Dekker.